## A small entertainment

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• A small entertainment: Three brothers are reunited celebrating the birthday of one of them. They realize that their ages in years are three different prime
Message 1 of 4 , May 7, 2006
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A small entertainment:

"Three brothers are reunited celebrating the birthday of one of them.
They realize that their ages in years are three different prime numbers.
This has happened eleven times in the course of its lives.
All of them has less than 100 years.

How many years old was the celebrated one when it happened the first time?

How many years old was the second (by age) brother when did born the third
one?

How many years old is the greater one?"

I apologize for the poor english ... Best regards,

Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosa@...
• ... time? ... third ... I think I have an answer -- scroll down to see. The only solution I could find was when we consider 1 to be a prime (which it not
Message 2 of 4 , May 8, 2006
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Ignacio wrote on 05/07/2006 12:37:08 PM:

> A small entertainment:
>
> "Three brothers are reunited celebrating the birthday of one of them.
> They realize that their ages in years are three different prime numbers.
> This has happened eleven times in the course of its lives.
> All of them has less than 100 years.
>
> How many years old was the celebrated one when it happened the first
time?
>
> How many years old was the second (by age) brother when did born the
third
> one?
>
> How many years old is the greater one?"
>
> I apologize for the poor english ... Best regards,
>

I think I have an answer -- scroll down to see.

The only solution I could find was when we consider 1 to be a
prime (which it not usually how it is defined). The answers are:

1) The oldest was 37 when it happened the first time.
2) The second-oldest was 6 when the third was born.
3) The oldest is 97.

I used a Pari program to generate pairs of differences in ages:
forprime(x=3,83,\
forprime(y=nextprime(x+1),89,\
forprime(z=nextprime(y+1),97,print(y-x," ",z-y))))

I took the output of this program, sorted the lines,
then counted the number of duplicates.
There were 10 each for the pairs of differences, "6,24" and "6,30".
If we allow 1 to be considered a prime, then the eleventh solution
comes from the second -- giving ages of 1,7 and 37.
The other 10 ages are:
5,11,41
7,13,43
11,17,47
17,23,53
23,29,59
31,37,67
37,43,73
47,53,83
53,59,89
61,67,97

[Non-text portions of this message have been removed]
• Monday, May 08, 2006 8:47 PM [GMT+1=CET], ... But 1 isn t prime ... ... / ... / ... / The three brothers hasn t the same birthday ... Best regards, Ignacio
Message 3 of 4 , May 8, 2006
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Monday, May 08, 2006 8:47 PM [GMT+1=CET],

> Ignacio wrote on 05/07/2006 12:37:08 PM:
>
>> A small entertainment:
>>
>> "Three brothers are reunited celebrating the birthday of one of them.
>> They realize that their ages in years are three different prime
>> numbers. This has happened eleven times in the course of its lives.
>> All of them has less than 100 years.
>>
>> How many years old was the celebrated one when it happened the first
>> time?
>>
>> How many years old was the second (by age) brother when did born the
>> third one?
>>
>> How many years old is the greater one?"
>>
>> I apologize for the poor english ... Best regards,
>>
>
> I think I have an answer -- scroll down to see.
>
> The only solution I could find was when we consider 1 to be a
> prime (which it not usually how it is defined). The answers are:

But 1 isn't prime ...

Hint:

|
|
\/

|
|
\/

|
|
\/

The three brothers hasn't the same birthday ...

Best regards,

Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosa@...
• ... You don t need to take 1 to be a prime to find a solution. The brother who is 1 year old reaches his 2nd birthday while the older two still have ages which
Message 4 of 4 , May 8, 2006
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>
> The only solution I could find was when we consider 1 to be a
> prime (which it not usually how it is defined)...

You don't need to take 1 to be a prime to find a solution.
The brother who is 1 year old reaches his 2nd birthday while
the older two still have ages which are odd primes.

Also, *both* of your progression patterns yield solutions to the
relative ages of the brothers.

This is a cute problem, in that there are two different "solutions"
to the relative ages of the brothers, but both solutions give the
same answers to the three questions...

The first time it happened, the "celebrated one" was celebrating
his 2nd birthday.

The second (by age) brother was five years old when the youngest
was born -- he was 7 years old on the 2nd birthday of the
youngest one.

The oldest brother's age at the time of the eleventh occurrence
is the same regardless of the pattern of their ages.

Cute problem, and deeper than it appears at first glance.

Jack
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